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 Question Electric field strength and potential gradient ( Physics Help and Math Help - Physics Forums Classical Physics )Updated: 2008-09-30 06:25:30 (7)
 Electric field strength and potential gradient A bit of a problem. My book teaches me that E = -(dV/dx), where E is the electric field strength, V is the electric potential, and x represents displacement. But, it also suggests along with the above formula that E = -(V/d) and displays a circuit with a battery of p.d. V and two parallel metal plates of distance (d) from each other. My question is, HOW did E = -(dV/dx) become E = -(V/d)??? The former formula, proven via differentiation, says that the electric field strength is negative of the potential gradient i.e. rate of change of electric potential with respect to the displacement. Then how does this transform into the electric field strength simply being equal to the negative of the ratio of the potential difference to the distance? It makes no sense to me! And yet, I've seen the latter formula being used in practice questions.
 Answers: Electric field strength and potential gradient ( Physics Help and Math Help - Physics Forums Classical Physics )
Electric field strength and potential gradient

 Originally Posted by jtbell The parallel-plate equation really should read to indicate that the voltage is the difference in potential between the two plates. Unfortunately it's a long-standing custom to use V to represent both potential (at a point) and potential difference (between two points) in different contexts.